Question
Show that the range of a bounded linear operator is not necessarily closed.
Hint: Use the linear bounded operator defined by .

Attempt to solution

My idea was to find an element that does not belong to the range and then try to build a convergent sequence in that has limit . The element satisfy the criteria because , with , but, clearly, , therefore, . The problem arise when I try to build the sequence, because with cannot converge to . Briefly, my problem is that I canīt find a limit point of that doesnīt belong to .

Jun 1st 2011, 10:25 PM

Drexel28

Quote:

Originally Posted by kinkong

Question
Show that the range of a bounded linear operator is not necessarily closed.
Hint: Use the linear bounded operator defined by .

Attempt to solution

My idea was to find an element that does not belong to the range and then try to build a convergent sequence in that has limit . The element satisfy the criteria because , with , but, clearly, , therefore, . The problem arise when I try to build the sequence, because with cannot converge to . Briefly, my problem is that I canīt find a limit point of that doesnīt belong to .

That's the right idea. How about something like defining the element of your sequence has in the first positions and zero afterwards-- i.e. . This is clearly in since it is equal to and it converges to

Jun 2nd 2011, 12:50 AM

Opalg

Quote:

Originally Posted by Drexel28

How about something like defining the element of your sequence has in the first positions and zero afterwards-- i.e. . This is clearly in since it is equal to and it converges to

The trouble there is that does not converge to in the space (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking Then is in since it is equal to , and it does converge to in the norm.

Jun 2nd 2011, 01:04 AM

Drexel28

Quote:

Originally Posted by Opalg

The trouble there is that does not converge to in the space (because the norm there is the sup norm). To avoid this difficulty, you could modify the construction slightly by taking Then is in since it is equal to ' and it does converge to in the norm.